Question

An alternating electric field, of frequency $\nu$, is applied across the Dees $($radius$=R)$ of a cyclotron that is being used to accelerate protons $($mass$=m,$ charge$=e)$. The operating magnetic field $\left(B\right)$ used in the cyclotron and the kinetic energy $\left(K\right)$ of the proton beam, produced by it, are given by

• A. $B=\frac{m\nu }{e}$ and $K=m^2\pi \nu R^2$
• B. $B=\frac{2\pi m\nu }{e}$ and $K=2m{\pi }^2{\nu }^2{R}^2$
• C. $B=\frac{m\nu }{e}$ and $K=2m{\pi }^2{\nu }^2{R}^2$
• D. $B=\frac{2\pi m\nu }{e}$ and $K=m^2\pi \nu R^2$

Answer 1

The correct option is B $B=\frac{2\pi m\nu }{e}$ and $K=2m{\pi }^2{\nu }^2{R}^2$

Frequency is given as $\left(\nu \right)=\frac{qB}{2\pi m}$

Where,
$q=$ charge on particle $=e$ (for proton)
$B=$ operating magnetic field
$m=$ mass of charged particle

$\therefore \nu =\frac{eB}{2\pi m}$ $\Rightarrow B=\frac{2\pi m\nu }{e}...\left(1\right)$


Also, $R=\frac{mv}{qB}$

Where,
$v=$ velocity of charged particle
$R=$ radius of circular path taken by charge

$\therefore v=\frac{qBR}{m}...\left(2\right)$

Now, kinetic energy, $K=\frac{1}{2}m{v}^2$

From equation $\left(2\right)$,

$K=\frac{1}{2}m\left[\frac{q^2{B}^2{R}^2}{m^2}\right]$

$\Rightarrow K=\frac{e^2{B}^2{R}^2}{2m}=\frac{e^2{R}^2}{2m}{\left[\frac{2\pi m\nu }{e}\right]}^2\left[\text{From (1)}\right]$

$\Rightarrow K=2m{\pi }^2{\nu }^2{R}^2$